Is I Irrational? by Colin Foster

Is i irrational? by Colin Foster I once saw a Key Stage 3 lesson in which pupils were Quite often, statements are not returned to and simplifying algebraic expressions, and where each reformulated as new things are learned. Assumptions question part (a, b, c, …) involved only the letter of that build up that may be helpful for a while but become question, so part b might have been something like problematic later (Tall, 2013). For example, typically 5b – 2b + b – 3b. When some of the pupils reached part i and were working on a question like 8i + 4i – 9i + 3i, numbers make an appearance. An irrational number is the idea of ‘irrational’ gets defined long before complexp the teacher stopped the class and suggested to the pupils any number which can’t be written in the form q , where (tongue in cheek, of course) that in that question part p and q are integers. Perhaps we should have said ‘any they were actually doing something much more advanced. real number …’? But then what would that have added for They could congratulate themselves that they were doing someone who hasn’t yet met any number that isn’t real, complex numbers, since in maths i actually means an or doesn’t even know that such a number could exist? It’s imaginary number (Note 1)! the sort of thing the teacher might slip in for the sake of their own mathematical conscience, but which is either Of course, the teacher was not being totally serious not noticed by the pupils or leads to questions which are about this, and had chosen to take an opportunity to introduce the pupils to the problem of square-rooting lesson. (If you are introducing irrational numbers today, a negative number, saying that we use the letter i as youlikely might to divert not want significantly today also from to be the the main lesson point where of youthe a ‘made-up number’ for − . But it got me thinking 1 have to introduce the idea of an imaginary number.) So whether in any sense it could be true to say that they one day, years later, when a pupil asks, “Is i irrational?” were ‘doing complex numbers’? On the face of it, their there may be a bit of work to do in sorting it out. activity didn’t look much different from that of a further It may be that this question arises partly because i is the numbers, simplifying expressions like 3 + 8i – 5 – 2i. Ismathematics i + i = 2i only student adding in theirimaginary first lessonnumbers on ifcomplex I think also be provoked by the cartoon (Fig. 1) that occasionally first letter of irrational (as well as imaginary). It might about the fact that i = −1 as I do it? Of course, it is only a convention to use i for −1; any other part of that the numbers i and , and has i saying to , “Be rational!” question could equally well have been thought of as anddoes the saying rounds to i “Getamong real!” sixth This formers, joke depends which personifies on the fact involving complex numbers, as of course nowhere did that is real but notπ rational, but mightπ be understood the book specify that the quantities were all real! as suggestingπ that i is rational but not real, otherwise i is beingπ hypocritical! So is i rational or irrational? Sometimes we do not always state clearly what we mean. For example, we might say that although we can factorize x2 – 1, as (x + 1)(x – 1), we can’t factorize x2 + 1, for example. However, just changing the letter name from x to z might suggest otherwise. On seeing z2 + 1, we might think in terms of working over the complex numbers, rather than over the reals, and so say that we could factorize it, as (z + i) (z – i) (Note 2). So the answer to “Is x2 + 1 irreducible?” depends on the set of numbers which we consider the choice of a letter x could push us towards considering onlycoefficients real numbers. to belong to, and something as slight as a Fig. 1 A number joke Mathematics in School, January 2018 The MA website www.m-a.org.uk 31 31-33-Foster QA is i irrational.indd 31 22/12/2017 09:14 … –3 < –2 < –1 < 0 < 1 < 2 < 3 … be expressed as the ratio of two integers. The only square Similarly, the imaginary axis is also a line, so presumably rootsBy our of definition integers whichabove, are i is rational not rational, are the since square it cannot roots we can write: of positive square numbers (and zero). Pupils might think that although –5 is not rational, –9 might possibly be, … –3i < –2i < –i < 0 < i < 2i < 3i … since 9 is a square number. However, writing = p , where –9 q But in general, the non-reals have two parts to them – q p2 = –9q2, which has no solutions for real p they are like ‘2-dimensional numbers’, so they can’t be and q, except p = q = 0, which contradicts the statement that put into order along a 1-dimensional line. q ≠ 0, leads to However, this is not right at all, as I only realized much So i is not rational — so is it irrational then? Not so fast! ≠ 0. So based on this we cannot call i rational. later on! The symmetry between the reals and the purely ‘Irrational’ is normally taken to refer to a real number imaginaries is deceptive. Maybe the use of vertical number which is not rational, so on this basis i is neither rational lines in school (e.g. temperature scales for working with nor irrational. Just as it doesn’t make sense to ask whether negative numbers) contributes to this easy false idea? But 2 is odd or even, because it is neither of the form 2n nor 3 the Argand diagram is not merely a horizontal number line of the form 2n + 1 (where n is an integer), it doesn’t make coupled with a vertical number line. The problem is that sense to ask whether i is rational or irrational. (Pupils we cannot say that 3i > 2i. This is quite counterintuitive: may think of ‘odd’ as any number that isn’t even but surely, whatever i is, 3 lots of it is more than 2 lots of it? But not state carefully that ‘number’ in this context means no! Let’s suppose that 3i > 2i. If we subtract 2i from both ‘integer’.) Rational and irrational are opposites, so it may sides, we get i > 0. But that cannot be true. If we multiply seem that a number which isn’t one must be the other, both sides of this inequality by i, since we are supposing and this is true across the reals, just as all integers are that i is positive (i > 0), the inequality sign will stay the either even or odd. However, the symmetry between same way round, giving i2 evens and odds is really quite unlike that between the know that i2 , giving us –1 > 0, which is clearly rationals and irrationals. We can specify what all the false. This means that we > must0. But haveby the been definition wrong of to i rationals are like (fractions with integer numerator and supposewe that i > 0,= or–1 that 3i > 2i. non-zero integer denominator), but the irrationals (of which, in a sense, there are far more) include not just the So we might think that, as it came out the wrong way surds but all kinds of strange things like e and , and even round at the end, with –1 > 0 instead of –1 < 0, we must lots of numbers which we can’t express in closed form. just have started the wrong way round at the beginning. The irrationals feel like the contents of a bin πinto which We have found that i isn’t positive, so it must be negative? we put all the badly-behaved numbers – and for this Let’s try that instead. Starting with i < 0, if we multiply reason it might feel like i really ought to go in that bin! both sides of this inequality by i, then since i is supposed to be negative now, this will reverse the inequality sign. Another problematic issue that arises with imaginary That means that we get i2 > 0 again, just like before, which numbers is to do with ordering. We know that for any we have just shown can’t be true! two real numbers, a and b, exactly one of the statements If i > 0 is false, and i < 0 is also false, perhaps we should a = b, a > b, a < b try i = 0? This seems unlikely. Starting with i = 0, we can must be true (the law of trichotomy). However, we can’t multiply both sides of this equation by i, and we get i2 = 0, do this with complex numbers. They can’t be ordered in which means –1 = 0, which is again false, of course. So we a useful way. If we take, for instance, 2 + 3i and 3 + 2i, we must have been wrong to say i = 0. can’t say This means that i is not greater than zero, or less than zero, 2 + 3i = 3 + 2i or equal to zero. We can’t use inequality signs between or 2 + 3i > 3 + 2i non-real numbers. I think I didn’t realize that this applied or 2 + 3i < 3 + 2i. to the purely imaginary numbers until at university I noticed that writing “Let > 0” also meant by implication None of those three statements is true.

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